How to create a new unknown or dynamic/expando object in Python

Question

In python how can we create a new object without having a predefined Class and later dynamically add properties to it ?

example:

dynamic_object = Dynami         


        

Answer 1

If you take metaclassing approach from @Martijn's answer, @Ned's answer can be rewritten shorter (though it's obviously less readable, but does the same thing).

obj = type('Expando', (object,), {})()
obj.foo = 71
obj.bar = 'World'

Or just, which does the same as above using dict argument:

obj = type('Expando', (object,), {'foo': 71, 'bar': 'World'})()

For Python 3, passing object to bases argument is not necessary (see type documentation).

But for simple cases instantiation doesn't have any benefit, so is okay to do:

ns = type('Expando', (object,), {'foo': 71, 'bar': 'World'})

At the same time, personally I prefer a plain class (i.e. without instantiation) for ad-hoc test configuration cases as simplest and readable:

class ns:
    foo = 71
    bar = 'World'

Update

In Python 3.3+ there is exactly what OP asks for, types.SimpleNamespace. It's just:

A simple object subclass that provides attribute access to its namespace, as well as a meaningful repr.

Unlike object, with SimpleNamespace you can add and remove attributes. If a SimpleNamespace object is initialized with keyword arguments, those are directly added to the underlying namespace.

import types
obj = types.SimpleNamespace()
obj.a = 123
print(obj.a) # 123
print(repr(obj)) # namespace(a=123)

However, in stdlib of both Python 2 and Python 3 there's argparse.Namespace, which has the same purpose:

Simple object for storing attributes.

Implements equality by attribute names and values, and provides a simple string representation.

import argparse
obj = argparse.Namespace()
obj.a = 123
print(obj.a) # 123 
print(repr(obj)) # Namespace(a=123)

Note that both can be initialised with keyword arguments:

types.SimpleNamespace(a = 'foo',b = 123)
argparse.Namespace(a = 'foo',b = 123)